Re: Peano's space-filling curve
From: KRamsay (kramsay_at_aol.com)
Date: 06/08/04
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Date: 08 Jun 2004 08:43:37 GMT
In article <2ij6g2Fmc61aU2@uni-berlin.de>,
"John Morgan" <john.morgan@REMOVECAPSataraxie.fr> writes:
|I have mentioned two points above. If you mean (1/3, 1/3)
|it's just what I said. But if you mean (1/3,2/3) you will
|need to explain to me why.
"Need"? Maybe you "need" to figure it out for yourself.
|The geometry screams out that it
|does not. Indeed every recursion of the Peano construct
|"creates" further points with rational coordinates that
|will never be crossed. Does the algorithm defining the
|drawing of the curve change somehow as one passes from the
|finite to the infinite?
This is an example of why proof is considered so important
in mathematics. The properties of the curve are typically
counter to people's intuitions. It's really your own
intuitions that are screaming, you see, but it's the intuitions
that are wrong, however commonsense they may seem.
Consider the definition of the curve given on this page:
http://www.cut-the-knot.org/Curriculum/Geometry/Peano.shtml
It seems Peano's original definition didn't involve a limiting
process, and makes it easier to see why each point of the square
is a point of the curve. His definition is less visual, and it
takes a little thinking to demonstrate that it's continuous, but
it's not so hard. It would be a good exercise for understanding
the whole thing to show that it's equivalent to the definition
in terms of successive approximations, and to prove it has the
claimed properties (being continuous and onto the whole square).
By that definition 5/36, 11/36, 13/36, and 19/36 all map
to (1/3,2/3). They have base 3 expansions
.01020202...
.02202020...
.10020202...
.11202020....
Take for example 5/36. It's in the second 1/9 of [0,1], i.e.
[1/9,2/9] (or [.01, .02] in base 3), which in the limit is being
mapped to the square 0<=x<=1/3, 1/3<=y<=2/3. But the interval
[11/81,12/81] (or [.0102,.011] in base 3) is mapped to the smaller
square 2/9<=x<=1/3, 5/9<=y<=2/3. The smaller interval
[101/729, 102/729] (or in base 3, [.010202, .01021]) is mapped
to the square 8/27<=x<=1/3, 17/27<=y<=2/3. And so on.
It's easier of course to see where the map sends values having
terminating base 3 expansions. But the successive approximations
send 5/36 to points whose distance from (1/3,2/3) goes to 0. And
*by definition* that means 5/36 gets sent to (1/3,2/3) in the
limiting curve.
|> The answers have been given many times, very clearly.
|> Why you're unable to glean them is hard to say.
|
|Simply, because your generalisation "very clearly" actually
|applies to the limited perspective that you and your fellow
|mathies have. I'm not going to re-iterate any of what I have
|said before about this. It's in Google.
You're trying to account for other people's greater ease in
learning this stuff by their having a more limited perspective
than yours? Or are you trying to say that we have too little
experience with what it takes to make things clear to the average
person?
We have collectively a lot of experience with students of
different aptitudes and attitudes. It's easy to see how big
a difference attitude can make. Many times students at various
levels have come to sci.math with the attitude that they (the
student) have something to teach those of us who know more
mathematics than they do, like the proper scientific attitude
to take, or what sort of mistakes they think we're making in
the mathematics itself. This second agenda, of trying to "sell"
some point they think they understand better than we do,
nearly always serves as a big distraction for the student.
Keith Ramsay
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